Why is bandwidth not centred at resonant frequency in LCR series circuit?

Question

When we calculate lower and higher frequency such that power becomes half in LCR circuit we end up with two value. For a complete derievation please see here. I have copied relevant information below:

1. Resonant frequency:

   \begin{align} X_L &= X_C \\ \omega_r L - \frac{1}{\omega_r C} &= 0 \\ \omega_r &= \frac{1}{\sqrt{LC}} \end{align}

2. Current: At frequency $\omega_r$, $Z_T$ is minimized and $I_S$ is maximized. \begin{align} I_\text{max} = \frac{V_\text{max}}{Z} = \frac{V_\text{max}}{\sqrt{R^2 + (X_L - X_C)^2}} = \frac{V_\text{max}}{\sqrt{R^2 + \left(\omega_r L - \frac{1}{\omega_r C} \right)^2}} \end{align}

3. Lower cut-off frequency $\omega_L$: At half power $P_m/2$, $I = I_m / \sqrt 2$ and $Z = \sqrt{2}R$. Then $X=-R$ (capacitive) and so $$\omega_L = -\frac{R}{2L} + \sqrt{\left(\frac{R}{2L}\right)^2 + \frac{1}{LC}}$$

4. Upper cut-off frequency $\omega_U$: Again $I = I_m/\sqrt{2}$ and $Z=\sqrt{2}R$, but this time $X=+R$ so $$\omega_U = +\frac{R}{2L} + \sqrt{\left(\frac{R}{2L}\right)^2 + \frac{1}{LC}}$$

5. Bandwidth: $$(\omega_L + \omega_U)/2 = \sqrt{\left( \frac{R}{2L}\right)^2 + \frac{1}{LC}}$$

Question: Does this mean bandwidth is not centered at the resonant frequency in an LCR series circuit?

Answer 1

Does this mean bandwidth is not centred at resonant frequency in LCR series circuit? If not, please explain.

It IS centered but on a logarithmic frequency axis, not a linear one. The center frequency is the geometric mean of the half-power frequencies, not the arithmetic mean

But we often write upper frequency = resonant frequency + R/2L and lower frequency = resonant frequency - R/2L

That's incorrect or, at best, an approximation for a relatively small bandwidth. We have

$$\frac{\omega_0}{\omega_{low}} = \frac{\omega_{high}}{\omega_{0}}$$ but $$\omega_{high} - \omega_{0} \neq \omega_0 - \omega_{low}$$

The half power frequencies are determined as the frequency where the reactance has the same magnitude as the resistance. Below the resonance, the impedance is dominated by the capacitor, i.e. the reactance is negative. Above the resonance its inductive, i.e. the reactance is positive.

We get a solution for each case

$$ \omega_{high} = \sqrt{\omega_0^2 + \frac{R^2}{4L^2}} + \frac{R}{2L} $$ $$ \omega_{low} = \sqrt{\omega_0^2 + \frac{R^2}{4L^2}} - \frac{R}{2L} $$

The bandwidth is the difference, which is $$ B = \omega_{high} - \omega_{low} = \frac{R}{L} $$

The product is $$ \omega_{high} \cdot \omega_{low} = \omega_0^2 + \frac{R^2}{4L^2} - \left( \frac{R}{2L} \right)^2 = \omega_0^2 $$

That means that resonance frequency is the geometric mean of the half-power frequencies.